How do you find the domain and range for $y=4x^2 - 2x$ ?

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Answer 1 · 2018-05-19T09:26:14

Domain: $(-oo, +oo)$ , Range: $[-1/4, +oo)$

$y = 4x^2-2x$

$y$ is defined $forall x in RR$

$:.$ the domain of $y$ is $(-oo,+oo)$

$y$ is a quadratic function of the form: $ax^2+bx+c$
Where: $a=4, b=-2, c=0$

Since $a>0$ , $y$ will have a minimum value where $x=-b/(2a)$

I.e. where $x= 2/(2xx4) = 1/4$

$:. y_min = y(1/4) = 4*(1/4)^2 - 2*(1/4)$

$= 1/4 - 1/2 =-1/4$

Since $y$ has no finite upper bound the range of $y$ is $[-1/4,+oo)$

We can infer these results from the graph of $y$ below.

graph{4x^2-2x [-2.35, 3.125, -0.713, 2.024]}

How do you find the domain and range for y=4x^2 - 2xy=4x^2 - 2x?